1. **Problem 1: Find the output $q$ that gives the highest revenue and compute the maximum revenue for** $$R(q) = -q^3 + 12q^2 - 36q + 200$$ 2. To find the maximum revenue, we first find the critical points by taking the derivative of $R(q)$ with respect to $q$ and setting it equal to zero. $$R'(q) = \frac{d}{dq}(-q^3 + 12q^2 - 36q + 200) = -3q^2 + 24q - 36$$ 3. Set the derivative equal to zero to find critical points: $$-3q^2 + 24q - 36 = 0$$ Divide both sides by $-3$: $$q^2 - 8q + 12 = 0$$ 4. Solve the quadratic equation using the quadratic formula: $$q = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 1 \times 12}}{2} = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2}$$ $$q = \frac{8 \pm 4}{2}$$ So, $$q_1 = \frac{8 + 4}{2} = 6, \quad q_2 = \frac{8 - 4}{2} = 2$$ 5. To determine which critical point is a maximum, evaluate the second derivative: $$R''(q) = \frac{d}{dq}(-3q^2 + 24q - 36) = -6q + 24$$ Evaluate at $q=6$: $$R''(6) = -6(6) + 24 = -36 + 24 = -12 < 0$$ Since $R''(6) < 0$, $q=6$ is a local maximum. Evaluate at $q=2$: $$R''(2) = -6(2) + 24 = -12 + 24 = 12 > 0$$ So $q=2$ is a local minimum. 6. Calculate the maximum revenue by substituting $q=6$ into $R(q)$: $$R(6) = -(6)^3 + 12(6)^2 - 36(6) + 200 = -216 + 432 - 216 + 200 = 200$$ --- 7. **Problem 2: Find the rate of change of the operating cost with respect to time $t$ when $t=4$ days for** $$C(t) = 5000 + 40 e^{0.2t} \sqrt{t + 3} + 15 \ln\left(10 + e^{0.3t}\right)$$ 8. Differentiate $C(t)$ with respect to $t$ term by term. - The derivative of the constant $5000$ is $0$. - For the second term $40 e^{0.2t} \sqrt{t + 3}$, use the product rule: Let $$u = e^{0.2t}, \quad v = \sqrt{t + 3} = (t+3)^{1/2}$$ Then $$u' = 0.2 e^{0.2t}, \quad v' = \frac{1}{2}(t+3)^{-1/2}$$ So, $$\frac{d}{dt}(uv) = u'v + uv' = 0.2 e^{0.2t} (t+3)^{1/2} + e^{0.2t} \times \frac{1}{2} (t+3)^{-1/2}$$ Multiply by 40: $$40 \times \left[0.2 e^{0.2t} (t+3)^{1/2} + e^{0.2t} \times \frac{1}{2} (t+3)^{-1/2}\right] = 8 e^{0.2t} (t+3)^{1/2} + 20 e^{0.2t} (t+3)^{-1/2}$$ - For the third term $15 \ln(10 + e^{0.3t})$, use the chain rule: $$\frac{d}{dt} \ln(f(t)) = \frac{f'(t)}{f(t)}$$ Here, $$f(t) = 10 + e^{0.3t}, \quad f'(t) = 0 + 0.3 e^{0.3t} = 0.3 e^{0.3t}$$ So, $$\frac{d}{dt} 15 \ln(10 + e^{0.3t}) = 15 \times \frac{0.3 e^{0.3t}}{10 + e^{0.3t}} = \frac{4.5 e^{0.3t}}{10 + e^{0.3t}}$$ 9. Combine all derivatives: $$C'(t) = 8 e^{0.2t} (t+3)^{1/2} + 20 e^{0.2t} (t+3)^{-1/2} + \frac{4.5 e^{0.3t}}{10 + e^{0.3t}}$$ 10. Evaluate $C'(t)$ at $t=4$: Calculate each part: - $e^{0.2 \times 4} = e^{0.8} \approx 2.2255$ - $(4 + 3)^{1/2} = 7^{1/2} \approx 2.6458$ - $(4 + 3)^{-1/2} = \frac{1}{2.6458} \approx 0.37796$ - $e^{0.3 \times 4} = e^{1.2} \approx 3.3201$ - $10 + e^{1.2} = 10 + 3.3201 = 13.3201$ Now substitute: $$C'(4) = 8 \times 2.2255 \times 2.6458 + 20 \times 2.2255 \times 0.37796 + \frac{4.5 \times 3.3201}{13.3201}$$ Calculate each term: - $8 \times 2.2255 \times 2.6458 \approx 47.12$ - $20 \times 2.2255 \times 0.37796 \approx 16.82$ - $\frac{4.5 \times 3.3201}{13.3201} \approx 1.12$ Sum: $$C'(4) \approx 47.12 + 16.82 + 1.12 = 65.06$$ **Final answers:** - The output $q$ that gives the highest revenue is $\boxed{6}$ units. - The maximum revenue is $\boxed{200}$. - The rate of change of the operating cost at $t=4$ days is approximately $\boxed{65.06}$ dollars per day.